Civil Engineering - Highway Engineering - Discussion
Discussion Forum : Highway Engineering - Section 1 (Q.No. 5)
5.
The total length of a valley formed by two gradients - 3% and + 2% curve between the two tangent points to provide a rate of change of centrifugal acceleration 0.6 m/sec2, for a design speed 100 km ph, is
Discussion:
30 comments Page 1 of 3.
VENKATESH said:
1 decade ago
Here N = 0.03+0.02 = 0.05.
L = 0.38(NV^3)^0.5.
L = 84.76MTS APPROX.
L = 0.38(NV^3)^0.5.
L = 84.76MTS APPROX.
Ram kumar said:
1 decade ago
Length is 84.16 means, then what s the answer? option A or option C.
Alik Biswas said:
1 decade ago
We Know, L= 2*(N*v^3/c)^0.5 (Length of valley curve).
Here, N = 2%-(-3%) (Algebraic difference in grades).
= 5% Or 0.05.
v = 100 km/hr or 27.778 m/sec (Design Speed).
c = 0.6 m/sec3 (Centrifugal acceleration).
Therefore, L= 2*(0.05*27.778^3/0.6)^0.5 = 84.53 m.
Say,84.60 m ([C] is the correct answer.
Here, N = 2%-(-3%) (Algebraic difference in grades).
= 5% Or 0.05.
v = 100 km/hr or 27.778 m/sec (Design Speed).
c = 0.6 m/sec3 (Centrifugal acceleration).
Therefore, L= 2*(0.05*27.778^3/0.6)^0.5 = 84.53 m.
Say,84.60 m ([C] is the correct answer.
Twizere kagunga elie said:
1 decade ago
I don't understand well if N is load or if is median.
Nitin said:
1 decade ago
N is sum of upward & downward gradient.
Manikanth said:
1 decade ago
Lenth of valley curve for comfort condition is 2(Nv^3/c)^0.5.
Tarun said:
1 decade ago
V = 1km/hr*5/18 is m/s.
Selvi said:
1 decade ago
L = 0.378(NV3)^1/2.
= 0.378(.05*100*100*100)^.5.
= 86.4 m.
= 0.378(.05*100*100*100)^.5.
= 86.4 m.
(1)
Rishabh said:
10 years ago
L is the length of valley curve.
And this formula is applicable in this case only 2(Nv^3/c)^0.5.
And this formula is applicable in this case only 2(Nv^3/c)^0.5.
Harpreet singh said:
10 years ago
I don't understand. Please explain.
(1)
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers